# proving a trig identity

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#1
prove the identity
[cos2A + sin2A - 1]/[cos2A - sin2A + 1] = tanA

i've been stuck on this one and i'm not sure where to start, some help would be great :-)
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1 month ago
#2
I would start by turning cos2A and sin2A into their cosA and sinA counterparts

Assuming you know the formulas?
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#3
(Original post by JustSomeGuy:/)
I would start by turning cos2A and sin2A into their cosA and sinA counterparts

Assuming you know the formulas?
yeah, the sin2A and cos2A identities.
that would be:

[cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]
Last edited by SashaNilssen; 1 month ago
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1 month ago
#4
(Original post by SashaNilssen)
yeah, the sin2A and cos2A identities.
that would be:

[cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]
Now I would divide the top and bottom expressions by cos^2A. It seems like a really random thing to do but you'll notice that now everything is in terms of tanA and secA, which is helpful because there's another formula that links them 2.
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#5
(Original post by JustSomeGuy:/)
Now I would divide the top and bottom expressions by cos^2A. It seems like a really random thing to do but you'll notice that now everything is in terms of tanA and secA, which is helpful because there's another formula that links them 2.
okay so after doing that i get

[cos^2A - sin^2A + sinA - 1] / [cos^2A- sin^A - sinA + 1]

hm, i don't see how this can be written with tanA and secA... did i do something wrong here?
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1 month ago
#6
(Original post by SashaNilssen)
okay so after doing that i get

[cos^2A - sin^2A + sinA - 1] / [cos^2A- sin^A - sinA + 1]

hm, i don't see how this can be written with tanA and secA... did i do something wrong here?
You didn't divide the WHOLE expression by cos^2A. So for example the top which is cos^2A-sin^2A+sinA-1 would become:

(cos^2A/cos^2A) - (sin^2A/cos^2A) + (sinAcosA/cos^A) - (1/cos^2A) and then simplify accordingly. You would do the same with the bottom as well.

The reason why is because if you divide the top AND bottom by cos^A it doesn't change the expression... it's like multiplying the fraction by 1 because you've done the same thing to the top and bottom so you have the same expression as before just it looks different so you can manipulate it better to get the answer you want. 0
#7
(Original post by JustSomeGuy:/)
You didn't divide the WHOLE expression by cos^2A. So for example the top which is cos^2A-sin^2A+sinA-1 would become:

(cos^2A/cos^2A) - (sin^2A/cos^2A) + (sinAcosA/cos^A) - (1/cos^2A) and then simplify accordingly. You would do the same with the bottom as well.

The reason why is because if you divide the top AND bottom by cos^A it doesn't change the expression... it's like multiplying the fraction by 1 because you've done the same thing to the top and bottom so you have the same expression as before just it looks different so you can manipulate it better to get the answer you want. okokok thats great, i simplified and i got to the final answer of tanA, but i have one more concern, what happens to the other numerator that was there before i divided by cos^2A?

i had [cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]

and then [cos^2A - sin^2A +sinAcosA - 1] / cos^2A

what happens to the original numerator?
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1 month ago
#8
(Original post by SashaNilssen)
okokok thats great, i simplified and i got to the final answer of tanA, but i have one more concern, what happens to the other numerator that was there before i divided by cos^2A?

i had [cos^2A - sin^2A +2sinAcosA - 1] / [cos^2 - sin^2A - 2sinAcosA + 1]

and then [cos^2A - sin^2A +sinAcosA - 1] / cos^2A

what happens to the original numerator?
Would you mind sending your working between how you got from the first thing to the second? If it helps you can just take a picture and send it from your phone.
Last edited by JustSomeGuy:/; 1 month ago
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#9
(Original post by JustSomeGuy:/)
Would you mind sending your working between how you got from the first thing to the second? If it helps you can just take a picture and send it from your phone.
i'm on my pc now but ill scribble the notes down quick and get the photo to my computer, give me a minute
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1 month ago
#10
In case I explained it bad this is what I meant by divide by cos^2A

Edit: sorry that it's sideways I didn't realize it would turn out that way lmao
Last edited by JustSomeGuy:/; 1 month ago
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#11
(Original post by JustSomeGuy:/)
In case I explained it bad this is what I meant by divide by cos^2A

Edit: sorry that it's sideways I didn't realize it would turn out that way lmao
ohhh okay i wont bother getting the image onto my pc, i understand it much better now thank you, i had the wrong idea
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1 month ago
#12
(Original post by SashaNilssen)
ohhh okay i wont bother getting the image onto my pc, i understand it much better now thank you, i had the wrong idea
Ah right cool my bad for the explanation then lol just tell me if you get it in the end or not and I'll be sure to help you out
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1 month ago
#13
(Original post by SashaNilssen)
okay so after doing that i get

[cos^2A - sin^2A + sinA - 1] / [cos^2A- sin^A - sinA + 1]

hm, i don't see how this can be written with tanA and secA... did i do something wrong here?
remember there are 3 versions of Cos(2a).

the one you chose is not the best for this question.
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#14
(Original post by JustSomeGuy:/)
Ah right cool my bad for the explanation then lol just tell me if you get it in the end or not and I'll be sure to help you out
wait no i have been mistaken yet again oops, i hit a wall
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#15
(Original post by the bear)
remember there are 3 versions of Cos(2a).

the one you chose is not the best for this question.
i considered that also but i didnt know which would be best
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1 month ago
#16
(Original post by SashaNilssen)
wait no i have been mistaken yet again oops, i hit a wall
My final tip to you is that sec^A = tan^A + 1 which is definitely a formula you should learn but you can derive it from sin^A+cos^A=1 if you divide both sides by cos^2A 0
1 month ago
#17
Referring to what "the bear" said. If you used cos2A = 1-2sin^2A for the top and cos2A=2cos^2A-1 for the bottom it would also give you the answer and would be seen as easier since you don't have the do the whole diving by cos^A thing however knowing which ones to use and where can be just as hard.
Last edited by JustSomeGuy:/; 1 month ago
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#18
(Original post by JustSomeGuy:/)
My final tip to you is that sec^A = tan^A + 1 which is definitely a formula you should learn but you can derive it from sin^A+cos^A=1 if you divide both sides by cos^2A yeah i know that one, i probably forgot, i did that and simplified and i got my answer.
thank you very much for your patience :-)
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1 month ago
#19
(Original post by SashaNilssen)
yeah i know that one, i probably forgot, i did that and simplified and i got my answer.
thank you very much for your patience :-)
It didn't even take that long but thank you! 0
1 month ago
#20
(Original post by SashaNilssen)
i considered that also but i didnt know which would be best
well if you see a + 1 or a - 1 as part of your expression, you should consider the version of Cos2A which has a -1 or a + 1 respectively
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